Question

The tangents to the parabola $y^2=4ax$ at $P\left(t_1\right)$ and $Q\left(t_2\right)$ intersect at $R$. Then the area of $△PQR$

• A. $\frac{a^2}{2}(t_1-t_2{)}^2$
• B. $\frac{a^2}{2}(|t_1-t_2|)$
• C. $\frac{a^2}{2}(|t_1-t_2|{)}^3$
• D. $a^2(t_1-t_2{)}^2$

Answer 1

The correct option is B $\frac{a^2}{2}(|t_1-t_2|)$

$y^2=4ax........\left(i\right)$

Tangents are drawn at $P\left(t_1\right)andQ\left(t_2\right)$

Intersects at $R$.

Points of intersection of tangents at $P\left(t^2\right)andQ\left(t^2\right)$

$=>R:({at}_1{t}_2,a(t_1+t_2\left)\right)$

Now tangent are drawn from $R$ to $equation\left(i\right)$

So, area of triangle formed by then tangent and chord of contact.

When tangents are drawn from $(x_1,y_1)=({y_1^2-4a{x}_1)}^{\frac{3}{2}}.\frac{1}{2a}$

Here, $x_1=a{t}_1{t}_2,y_1=a(t_1+t_2)$

$=>Areaof△RPQ=\frac{1}{2a}\left[\right(a(t_1+t_2{]}^2-4a({a{t}_1{t}_2\left)\right]}^{\frac{3}{2}}$

$=\frac{1}{2a}\left({a^2)}^{\frac{3}{2}}\right[({{t_1+t_2)}^2-4t_1{t}_2]}^{\frac{3}{2}}$

$=\frac{a^2}{2}\left({{t_1-t_2)}^2]}^{\frac{3}{2}}\right(\therefore a^2+b^2-2ab=({a^2+b^2)}^2-4ab)$

$=\frac{a^2}{2}[|t_1-t_2|{]}^3$